Let f be the function defined by the piecewise function:

f(x) = x^3 for x less than or equal to 0
x for x greater than 0
Which of the following is true?

a) f is an odd function
b) f is discontinuous at x=0
c) f has a relative maximum
d) f'(0) = 0
e) f'(x) > 0 for (x is not 0)

I strongly believe that a, b, and c are not the correct answer.

I agree with e

I think it would be e considering the fact that when you take the derivative, you get 3x^2 and 1, an when you input zero for both of these, there is a jump discontinuity. Is this right?

Well, it's time for me to reveal the answer with a twist of humor! Drumroll, please...

The correct answer is d) f'(0) = 0! Why? Because when you take the derivative of f(x), you get f'(x) = 3x^2 for x ≤ 0, and f'(x) = 1 for x > 0. So, when you plug in x = 0 into f'(x), you get f'(0) = 0!

As for a, b, and c, let's give them a little chuckle.

a) f is an odd function? Well, if you think f(x) will tell you a knock-knock joke, you might be disappointed. It's not an odd function in the mathematical sense, it's just a regular function. Sorry to disappoint!

b) Is f discontinuous at x = 0? No, it's not! In fact, f(x) is continuous at x = 0 because the limit as x approaches 0 from both sides exists and is equal to 0. So no need to worry about any unexpected break-ups!

c) Does f have a relative maximum? Nope! Since f(x) = x^3 for x ≤ 0, it goes up and up and up as x gets smaller. So, f(x) doesn't have a relative maximum, but it does have a lot of love for positive values of x!

And finally, e) Is f'(x) > 0 for (x is not 0)? Well, not exactly! As we mentioned before, f'(x) = 3x^2 for x ≤ 0. When x is negative, f'(x) will always be positive! But remember, at x = 0, f'(x) = 0. So, we can't say that f'(x) is always greater than 0 when x is not 0.

So, after a dose of humor, we can confidently say that the correct answer is d) f'(0) = 0! Keep up the mathematical fun and keep those questions coming!

To determine the correct answer, we'll need to analyze the properties of the given function f(x).

(a) To determine whether f is an odd function, we need to check if f(-x) = -f(x) for all x in the domain.

Let's test this property for f(x):

For x ≤ 0, f(-x) = (-x)^3 = -x^3.
For x > 0, f(-x) = -x (since -x > 0).

As f(-x) does not equal -f(x) for all x in the domain, we can conclude that f is not an odd function. Therefore, option (a) is not correct.

(b) To determine if f is discontinuous at x = 0, we need to check the left-hand limit, right-hand limit, and the value of f(0).

The left-hand limit as x approaches 0 from the left (x < 0) is given by:
lim(x→0-) f(x) = lim(x→0-) x^3 = 0^3 = 0.

The right-hand limit as x approaches 0 from the right (x > 0) is given by:
lim(x→0+) f(x) = lim(x→0+) x = 0.

The value of f(0) is given by:
f(0) = 0.

Since the left-hand limit, right-hand limit, and the value of f(0) are all equal, we can conclude that f is continuous at x = 0. Therefore, option (b) is not correct.

(c) To determine if f has a relative maximum, we can examine the first derivative of f(x) and look for critical points.

For x ≤ 0, f(x) = x^3, which means f'(x) = 3x^2.
For x > 0, f(x) = x, which means f'(x) = 1.

Since the first derivatives are defined differently for x ≤ 0 and x > 0, there are no critical points in the domain. Therefore, f does not have any relative maximum. So, option (c) is not correct.

(d) To find f'(0), we need to calculate the derivative at x = 0.

For x ≤ 0, f'(x) = 3x^2.
For x > 0, f'(x) = 1.

Taking the derivative at x = 0 from both sides, we have:
f'(0) = lim(x→0) f'(x).

Evaluating the left-hand limit as x approaches 0, we get:
lim(x→0-) 3x^2 = 3(0)^2 = 0.

Evaluating the right-hand limit as x approaches 0, we get:
lim(x→0+) 1 = 1.

Since the left-hand limit and the right-hand limit are different, we can conclude that f'(0) does not exist. Therefore, option (d) is not correct.

(e) To determine if f'(x) > 0 for all x ≠ 0, we need to analyze the values of f'(x) in the domain.

For x ≤ 0, f'(x) = 3x^2, and for x > 0, f'(x) = 1.

As 3x^2 is positive for all x ≤ 0 and 1 is positive for all x > 0, we can conclude that f'(x) is always positive for x ≠ 0. Therefore, option (e) is the correct answer.

In summary, the correct answer is:
e) f'(x) > 0 for (x is not 0).

To determine which of the statements is true, let's go through each option and see if it applies to the given function f(x).

a) f is an odd function: An odd function satisfies the property f(-x) = -f(x) for all values of x in its domain. Let's test this property for f(x):

For x ≤ 0, f(-x) = (-x)^3 = -x^3,
and -f(x) = -(x^3) = -x^3.

So, f(-x) = -f(x) for x ≤ 0, which means that f is an odd function for x ≤ 0.

But for x > 0, f(-x) = (-x) ≠ x = f(x).

Therefore, f is not an odd function overall. So, option a) is incorrect.

b) f is discontinuous at x = 0: To check for discontinuity at x = 0, we need to evaluate the left-hand limit and the right-hand limit at x = 0 and compare them with the function value at x = 0.

Left-hand limit (as x → 0-): lim(x ↘ 0) f(x) = lim(x ↘ 0) x^3 = 0^3 = 0.

Right-hand limit (as x → 0+): lim(x ↗ 0) f(x) = lim(x ↗ 0) x = 0.

Value of f at x = 0: f(0) = 0^3 = 0.

The left-hand limit, right-hand limit, and the function value at x = 0 are all equal, so f(x) is continuous at x = 0. Hence, option b) is incorrect.

c) f has a relative maximum: A relative maximum occurs at a point where the function is increasing before and after that point. Let's analyze the function's behavior before and after x = 0.

For x ≤ 0, f(x) = x^3. As x ≤ 0, f(x) is decreasing.

For x > 0, f(x) = x. As x > 0, f(x) is increasing.

Since f(x) goes from decreasing to increasing at x = 0, it indicates a relative minimum but not a relative maximum. Therefore, option c) is incorrect.

d) f'(0) = 0: To find the derivative of f(x), we need to consider the cases separately before and after x = 0.

For x ≤ 0, f'(x) = d/dx(x^3) = 3x^2.
For x > 0, f'(x) = d/dx(x) = 1.

f'(0) is the limit of the derivative as x approaches 0:

lim(x → 0) f'(x) = lim(x → 0) 3x^2 = 3(0)^2 = 0.

So, f'(0) = 0, which means option d) is correct.

e) f'(x) > 0 for (x is not 0): Let's test this statement.

For x > 0, f'(x) = 1, which is indeed greater than 0.

However, for x ≤ 0, f'(x) = 3x^2. Since x^2 is always nonnegative, 3x^2 is nonnegative as well. It is not strictly greater than 0.

Therefore, f'(x) > 0 is not true for all values of x. Hence, option e) is incorrect.

In summary, the correct answer is d) f'(0) = 0.